Question

3. 18 points] Warm-blooded animals use large quantities of energy to maintain body temperature because of heat loss through the body surface. In fact, biologists believe that the primary energy drain on a resting warm-blooded energy available is proportional to the blood flow through the lungs, which is circulated, trhe amount of available energy will equal the amount of energy weights of some mammals to their heart rate in beats per minute. Based on weight. Discuss the assumptions of your model. Use the data to check your animal is the maintenance of body temperature. Assume the amount of the source of oxygen. Assuming the least amount of blood needed is used to maintain the body temperature. The following data relates the the previous discussion construct a model that relates heart rate to body model Mammal Mouse Guinea Pig Rabbit Little Dog Sheep Horse Body weight(g) 25 300 2000 5000 50,000 450,000 Pulse Rate 670 300 205 120 70 38

Answer 1

Assumptions It is well-known that small warm-blooded animals have much higher resting pulse rates than large ones. For examples, a canary, with a body mass of 20 grams, may have a heart rate of 1000 beats/minute, as compared to 70-80 beats/minute for a human being Warm-blooded animals at rest use large quantities of energy to maintain body temperature, due to heat loss through the body surface. Let us make the following assumptions about energy balance over a fixed time interval (1 minute, say) A. Energy lost is proportional to the surface area of the body B. Energy gained is proportional to the amount of blood flowing through the lungs, the source of oxygen for the body C. Energy lost - energy gained for a body in temperature equilibrium. Model Construction E- energy S surface area m mass W- weight of the animal r = pulse rate V volume of the animal bvolume of blood pumped by the heart in one stroke From assumption A, we can assume that the energy required to maintain the body temperature of a bird E is proportional to its body surface area S, i.e. E ox S Since blood flow through the lungs provides animals oxygen, larger animals have larger lungs and require more oxygen. Therefore, we assume that the amount of energy available is proportional to the blood flow b through the lungs, i.e E ox b Also, we assume that the least amount of blood needed to circulate, the amount of available energy will equal the amount of energy used to maintain the body temperature. This implies b x S Since larger animals have larger hearts and larger lungs, we can assume the birds are geometrically similar. Thus, S volume of the animal V-where V is the Since V m x W where m is the mass of the animal and W is the weight of the animal, we have S Wī . This implies b oc Wi Also, since blood flow is determined by the total amount of blood pumped by the heart per unit time (minutes), and every time the heart beats the blood is pumped into the arteries. We have b OCTV oc rW where r is the pulse rate and vh is heart volume. From b W-and b。CTW, we Methodology .First we will be modeling using Geometric Similarity given our power curve model function ·We will be using Least squares criterion to fit our curve model of the form y = Azn where n is fixed to a given collection of data points. In our case, our curve is r-AWT. Therefore we will use the least squares criterion to estimate A with n-_1/3 .In addition we will use the Chebyshev criterion to find the bounds

body_weight<- c(25,300,2000,5000,50000,450000)
pulse_rate<-c(670,300,205,120,70,38)

y <- pulse_rate
x <- body_weight^(-1/3)
model <- lm (y ~x)
summary(model)

Call:
lm(formula = y ~ x)

Residuals:
      1       2       3       4       5       6 
 -2.238  -5.771  32.430 -12.826  -3.206  -8.390 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)    21.56      10.53   2.048     0.11    
x            1902.59      66.62  28.560 8.94e-06 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 18.27 on 4 degrees of freedom
Multiple R-squared:  0.9951,    Adjusted R-squared:  0.9939 
F-statistic: 815.7 on 1 and 4 DF,  p-value: 8.945e-06

R^2 = 0.9951 which is quite high

hence this is good model

r = 21.56 + 1902.59 (W)^-1/3